2

I know that according to the Fundamental theorem of linear algebra the row space and the null space are orthogonal, but I don't really understand why. Could someone give an intuitive explanation of why this is with maybe some examples from $\mathbb{R}^2$ or $\mathbb{R}^3$ with the standard Euclidean inner product?

Robert S. Barnes
  • 2,427
  • It's just how matrix multiplication works --- when you multiply a matrix $A$ times a vector $v$, the $j$th entry in the product is the inner product of the $j$th row of $A$ with $v$. So $v$ is in the nullspace if and only if it's orthogonal to the generators of the row space. – Gerry Myerson Oct 14 '12 at 08:46

1 Answers1

4

The row space is the set of $A^Tx$ for every vector $x$, the null space is the set of vectors $y$ such that $Ay=0$. The scalar product between a vector in the row space and a vector in the null space is $\langle y,A^Tx\rangle=y^T(A^Tx)=x^T(Ay)=x^T0=0$. The second equality follows from the fact that $y^TA^Tx$ has size $1\times 1$, hence is equal to its transpose $x^TAy$.

Did
  • 279,727
  • 1
    And the first equality follows from the fact that $\langle u,v\rangle = u^Tv$ – Robert S. Barnes Oct 15 '12 at 08:55
  • 1
    Couldn't you have written directly without resorting to the $1\times 1$: $\langle y,A^Tx\rangle=(A^Tx)^Ty=(x^TA)y=x^T(Ay)=x^T0=0$ since $\langle u,v\rangle = \langle v,u\rangle$? – Robert S. Barnes Oct 15 '12 at 09:00
  • Sure. And $\langle u,v\rangle=\langle v,u\rangle$ because this is the $1\times1$ case... – Did Oct 15 '12 at 09:54